They are the triangles that contain a 90 degree angle and two acute angles
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If you draw a line perpendicular to side c and intersecting the vertex of angle C, you make two right triangles. Let's call the perpendicular segment side d. Now the sinB = d/a, so d = a sinB = 24 * sin87° ≈ 23.967. SinA = d/b, so b = d ÷ sinA = 23.967 ÷ sin42° ≈ 36.
Let ABC and DEF be triangles which are right angled at A and D, such that the hypotenuses BC and EF are equal and, without loss of generality, angle B = angle E.ThenThen by the sine rule, BC/sin(A) = AC/sin(B) and EF/sin(D) = DF/sin(E)Since angle A = angle D = pi/2 radians, then sin(A) = sin(D) = 1so that BC/sin(A) = BC while EF/sin(D) = EFtherefore, since the hypotenuses BC and EF are equal, the left hand sides of the two equations are equal.Therefore, AC/sin(B) = DF/sin(E)then, since angle B = angle E, then sin(B) = sin(E) so that AC = DF.Also, angle C = pi/2 - angle Band angle F = pi/2 - angle Ethe right hand sides are equal so angle C = angle F.Then in a manner similar to the above, we can show that AB = DE.Thus all three pairs of corresponding sides are equal and all three pairs of corresponding angles are equal so that the two triangles are congruent.
LEG!
A circle, it has no angles.There are a infinite number of 2-d shapes that have no right angles.TrapeziumRhombusAll polygons except a square or a rectangleAll triangles except a right angle triangleAny irregular shapeetc.etc.
First identify the two lines that are parallel. Call these the top and bottom lines. Mark the two corners on the left A and B. Then angle A + angle B = 180 degrees. Next, mark the two corners on the right C and D. Angle C + angle D = 180 degrees. Hope this helps :)